Question

Stability of compounds in new drugs. Refer to the ACS Medicinal Chemistry Letters (Vol. 1, 2010) study of the metabolic stability of drugs, Exercise 2.22 (p. 83). Recall that two important values computed from the testing phase are the fraction of compound unbound to plasma (fup) and the fraction of compound unbound to microsomes (fumic). A key formula for assessing stability assumes that the fup/fumic ratio is 1:1. Pharmacologists at Pfizer Global Research and Development tested 416 drugs and reported the fup/fumic ratio for each. These data are saved in the FUP file, and summary statistics are provided in the accompanying Minitab printout. Suppose the pharmacologists want to determine if the true mean ratio, μ, differs from 1.

a. Specify the null and alternative hypotheses for this test.

b. Descriptive statistics for the sample ratios are provided in the Minitab printout on page 410. Note that the sample mean,\(\overline x = .327\)is less than 1. Consequently, a pharmacologist wants to reject the null hypothesis. What are the problems with using such a decision rule?

c. Locate values of the test statistic and corresponding p-value on the printout.

d. Select a value of\(\alpha \)the probability of a Type I error. Interpret this value in the words of the problem.

e. Give the appropriate conclusion based on the results of parts c and d.

f. What conditions must be satisfied for the test results to be valid?

Short answer

a. The null and alternative hypotheses are\({H_0}:\mu = 1\)and\({H_a}:\mu \ne 1\)

b.In this situation, the examiner must observe if the sample means the value is unusual when the true mean ratio is 1.

c.From the MINITAB output, the test statistic values are\(z = - 47.09\), and the p-value is 0.000.

d. It can conclude that the probability of the true mean ratio differs from 1, but in reality, the true mean ratio is 1.

e. There is enough evidence to infer that the true mean ratio differs from 1.

f. The distribution of the sample mean\(\left( {\overline x } \right)\) is approximately normal without considering the population distribution and assumes that the sample is a simple random sample.

Step-by-step solution

Given information

MINITAB output is as follows:

Specifying the null and the alternative hypothesis

a.

Let \(\mu \) be the true mean ratio.

Null hypothesis:

\({H_0}:\mu = 1\)

That is, the true mean ratio is 1.

Alternative hypothesis:

\({H_a}:\mu \ne 1\)

That is, the true mean ratio differs from 1.

Explaining the problems

b.

The sample mean does not make any changes in the sampling process. In this situation, the examiner must observe if the sample mean value is unusual when the true mean ratio is 1.

Interpretation of the given output

c.

From the MINITAB output, the test statistic values are \(z = - 47.09\), and the p-value is 0.000.

Interpretation for Type I error

d.

Type I error is the error committed to rejecting a null hypothesis when it is true. The conclusion is that the true mean ratio differs from 1, but the true mean ratio is 1.

Choose a value of 0.05. Then the probability of error committed to rejecting a null hypothesis when it is true is 0.05. Thus, we can conclude that the probability of the true mean ratio differs from 1, but in reality, the true mean ratio is 1.

Interpretation for Type I error

e. If a p-value is less than \(\alpha \), then the null hypothesis is rejected. Here, the p-value is 0.000, which is lesser than the level of significance. That is, the p-value \(\left( {p = 0.000} \right) < \left( {\alpha = 0.05} \right)\)

Therefore, reject the null hypothesis at 0.05 level of significance.

Thus, there is enough evidence to infer that the true mean ratio differs from 1.

Conditions for validating the results

f.

The given sample size is 416, which is greater than 30. Thus, the distribution of the sample mean\(\left( {\overline x } \right)\) is approximately normal without considering the population distribution and assumes that the sample is a simple random sample. Therefore, the Central limit theorem is appropriate for the given data.